Find NCERT Exemplar Solutions for all the questions of Class 12 Physics Chapter 10 – Wave Optics here. These questions are very important for competitive exams like IIT JEE (JEE Main and JEE Advanced), UPSEE, WBJEE, VITEEE, SRMJEEE and NEET etc., as the difficulty level of these questions is quite higher. These solutions are explained by experienced Subject Experts of Physics.

Important Topics of Chapter 10 – Waves Optics:

Huygen's principle,

Reflection and refraction of plane wave at a plane surface using wave fronts,

In a Young's double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case,

(a) There shall be alternate interference patterns of red and blue

(b) There shall be an interference pattern for red distinct from that for blue

(c) There shall be no interference fringes

(d) There shall be an interference pattern for red mixing with one for blue

Solution:(c)

We know that, for the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength.

In a Young's double-slit experiment, if one of the holes is covered by a red filter and another by a blue filter, then only red and blue lights are present due to alteration. In Young's double-slit experiment, a monochromatic light is used for the formation of fringes on the screen.

Therefore, there shall be no interference fringes.

Question:

Consider sunlight incident on a pinhole of width 103 Å. The image of the pinhole seen on a screen shall be

(a) A sharp white ring

(b) Different from a geometrical image

(c) A diffused central spot, white in colour

(d) Diffused coloured region around a sharp central white spot

Solution: (b, d)

We are given that, the width of pinhole is 103 Å or 1000 Å

We know that, the wavelength of visible sunlight lies between 4000 Å and 8000 Å.

Since, the wavelength (l) is less than the width of the slit.

Hence, the light is diffracted from the hole.

Due to diffraction the image formed on the screen will be different from the geometrical image.

Question:

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased

(a) The size decreases

(b) The intensity increases

(c) The size increases

(d) The intensity decreases

Solution: (a, b)

(a) The width of central maximum of diffraction pattern of hole is inversely proportional to the size of the hole.

Therefore, the width (size) of central maximum decreases when size of the hole increases.

(b) Since, the amount of light energy distributed over a small area is same.

Hence, the intensity will increase.

Question:

Why is the diffraction of sound waves more evident in daily experience than that of light wave?

Solution:

The diffraction occurs only if the wavelength of waves and width of the slit is almost equal.

But we know that, the frequencies of sound waves lie between 20 Hz and 20000 Hz. Therefore, the range of their wavelength is between 15 m and 15mm.

Now, the wavelength of light waves is 7000 × 10-10 m to 4000 × 10-10 m. The width of the slit is very near to the wavelength of sound waves as compared to light waves. This is the reason why the diffraction of sound waves is more evident in daily life than that of light waves.